When you have a set of two equations, you need to solve one for one of the variables. Then, you substitute that value into the other. Either equation will work rather well, so use the first. Thus, from  you get:

Now, substitute into the second equation:

Simplifying, you get:

Begin by writing out your data in equations. Based on the description, we know:

and ...

Now, to solve two equations like this, solve one equation for a variable. Then substitute into the other. The easiest one to solve is . From this, we know:

Now, substitute this into the other equation:

Simplify:

Luckily, this is just what we need!

There are many methods to finding the solution of a system of linear equations: graphing, elimination, substitution, matrix manipulations; we'll use subsitution.

Since both equations are equal to , we can use substitution to set them equal to each other and solve for x:

Now we plug the value we just found for  into either equation (it doesn't matter, check both if you're unsure)

Thus we get the ordered pair: 

The easiest way to solve this equation is to work backwards. Start with the  for the third year and divide by  to get the second year's earning which were , then subtract  to get the first year which was .

You factor the equation and get:

(x + 2)(x + 1) = 0

Then, you get x = –2 and x = –1.

There are two ways to solve this problem. One option is to use the quadratic formula:

In this problem, a = 1, b = 1, and c = –6. However, there is a lot of room for computational error using this method.

Another option is to factor the equation using the reverse FOIL method. Because we have x², we know that both factors must include x. Thus, so far we have (x   )(x   ).  We also know that one factor must be negative and one must be positive if our last value (the L in FOIL) is to be negative. Thus, we have (x–  ) (x+ ). The factors of 6 are 1, 2, 3, and 6, so our last values must multiply together to make 6. Therefore, our options are (x – 1)(x + 6), (x – 6)(x + 1), (x – 2)(x + 3), and (x – 3)(x + 2). However, because we need our middle value to be 1, the combination of the outer and inner values (O and I in FOIL), must add up to positive 1. Since only a combination of +/–2 and +/–3 can add to 1, we can eliminate our first two options. Then, because we need the positive value to be larger than the negative value, we can determine that the correct factorization is (x – 2)(x + 3). When we set each of these pieces equal to zero, we get x – 2 = 0 and x + 3 = 0. Thus, X equals 2 and –3. We can check our work using FOIL.

1. Factor the above equation:

2. Solve for :

and...

To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

(x + 2)(x + 3) = 0

x = –2 or x = –3

To solve, begin by factoring the equation. We know that in our two factors, one will begin with  and one will begin with . Fiddling with different factors of  (we are looking for two numbers that, when multiplied by  and  separately, will add to ), we come to the following:

(If you are unsure, double check by expanding the equation to match the original)

Now, set the each factor equal to 0:

Do the same for the second factor:

Therefore, our two values of  are  and .

Alternatively, this problem can be solved by plugging each answer choice into the original equation and finding which set of numbers make the equation equate to 0.

Given a quadratic equation equal to zero you can factor the equation and set each factor equal to zero. To factor you have to find two numbers that multiply to make 9 and add to make 6. The number is 3. So the factored form of the problem is (x+3)(x+3)=0. This statement is true only when x+3=0. Solving for x gives x=-3. Since this problem is multiple choice, you could also plug the given answers into the equation to see which one works.